It’s -7, not -3 because cd is 16 apart and 1/8 is 2 units to the right of c making it -7.
Answer:
<em>Both buckets will have the same amount of water after 25 minutes</em>
Step-by-step explanation:
<u>Equations</u>
Let's call:
t = number of minutes elapsed for both buckets
The red bucket already has 200 L of water and it's flowing in at a rate of 6 liters per minute. Thus, the volume of water in this bucket is:
R = 200 + 6t
The blue bucket has 500 liters of water and is being drained at 6 liters per minute. The volume can be modeled by the equation:
B = 500 - 6t
The condition for both buckets to have the same amount of water is:
R = B
200 + 6t = 500 - 6t
Adding 6t and subtracting 200:
6t + 6t = 500 - 200
12t = 300
t = 300/12
t = 25 minutes
Both buckets will have the same amount of water after 25 minutes
Answer:
![A:\\R=4\hspace{8}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z](https://tex.z-dn.net/?f=A%3A%5C%5CR%3D4%5Chspace%7B8%7DG%3D%5C%7By%5Chspace%7B3%7D%20%3B%5Chspace%7B3%7D1%5Cleq%20y%5Cleq%206%20%5C%7D%5Chspace%7B3%7Dy%5Cin%20Z)
![B:\\R+G=5\hspace{8}R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{5}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{3}and\hspace{3}R\neq G](https://tex.z-dn.net/?f=B%3A%5C%5CR%2BG%3D5%5Chspace%7B8%7DR%3D%5C%7Bx%5Chspace%7B3%7D%20%3B%5Chspace%7B3%7D1%5Cleq%20x%20%5Cleq%206%20%5C%7D%5Chspace%7B3%7Dx%5Cin%20Z%5Chspace%7B5%7DG%3D%5C%7By%5Chspace%7B3%7D%20%3B%5Chspace%7B3%7D1%5Cleq%20y%5Cleq%206%20%5C%7D%5Chspace%7B3%7Dy%5Cin%20Z%5Chspace%7B3%7Dand%5Chspace%7B3%7DR%5Cneq%20G)
![C:\\R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{3}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{5}R =3 \hspace{3}or\hspace{3}G=3](https://tex.z-dn.net/?f=C%3A%5C%5CR%3D%5C%7Bx%5Chspace%7B3%7D%20%3B%5Chspace%7B3%7D1%5Cleq%20x%20%5Cleq%206%20%5C%7D%5Chspace%7B3%7Dx%5Cin%20Z%5Chspace%7B3%7DG%3D%5C%7By%5Chspace%7B3%7D%20%3B%5Chspace%7B3%7D1%5Cleq%20y%5Cleq%206%20%5C%7D%5Chspace%7B3%7Dy%5Cin%20Z%5Chspace%7B5%7DR%20%3D3%20%5Chspace%7B3%7Dor%5Chspace%7B3%7DG%3D3)
D:
![R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{3}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{5};\hspace{3}R+G\neq8](https://tex.z-dn.net/?f=R%3D%5C%7Bx%5Chspace%7B3%7D%20%3B%5Chspace%7B3%7D1%5Cleq%20x%20%5Cleq%206%20%5C%7D%5Chspace%7B3%7Dx%5Cin%20Z%5Chspace%7B3%7DG%3D%5C%7By%5Chspace%7B3%7D%20%3B%5Chspace%7B3%7D1%5Cleq%20y%5Cleq%206%20%5C%7D%5Chspace%7B3%7Dy%5Cin%20Z%5Chspace%7B5%7D%3B%5Chspace%7B3%7DR%2BG%5Cneq8)
Step-by-step explanation:
Let:
![R=The\hspace{3}number\hspace{3}shows\hspace{3}by\hspace{3}the\hspace{3}red\hspace{3}dice\\G=The\hspace{3}number\hspace{3}shows\hspace{3}by\hspace{3}the\hspace{3}green\hspace{3}dice](https://tex.z-dn.net/?f=R%3DThe%5Chspace%7B3%7Dnumber%5Chspace%7B3%7Dshows%5Chspace%7B3%7Dby%5Chspace%7B3%7Dthe%5Chspace%7B3%7Dred%5Chspace%7B3%7Ddice%5C%5CG%3DThe%5Chspace%7B3%7Dnumber%5Chspace%7B3%7Dshows%5Chspace%7B3%7Dby%5Chspace%7B3%7Dthe%5Chspace%7B3%7Dgreen%5Chspace%7B3%7Ddice)
A:
This is easy, simply R=4, and we don't know the value of G, so:
![R=4\hspace{8}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z](https://tex.z-dn.net/?f=R%3D4%5Chspace%7B8%7DG%3D%5C%7By%5Chspace%7B3%7D%20%3B%5Chspace%7B3%7D1%5Cleq%20y%5Cleq%206%20%5C%7D%5Chspace%7B3%7Dy%5Cin%20Z)
B: Be careful here, we know the numbers add to 5, so we don't know the exact value of R or G, because R could be 4 and G could be 1 or R could be 2 and green could be 3. However we can be sure that they can´t have the same value, because 5 is an odd number, if we add the same number to a number the result is always even: n+n=2n. So:
![R+G=5\hspace{8}R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{5}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{3}and\hspace{3}R\neq G](https://tex.z-dn.net/?f=R%2BG%3D5%5Chspace%7B8%7DR%3D%5C%7Bx%5Chspace%7B3%7D%20%3B%5Chspace%7B3%7D1%5Cleq%20x%20%5Cleq%206%20%5C%7D%5Chspace%7B3%7Dx%5Cin%20Z%5Chspace%7B5%7DG%3D%5C%7By%5Chspace%7B3%7D%20%3B%5Chspace%7B3%7D1%5Cleq%20y%5Cleq%206%20%5C%7D%5Chspace%7B3%7Dy%5Cin%20Z%5Chspace%7B3%7Dand%5Chspace%7B3%7DR%5Cneq%20G)
C: R is 3 or G is 3, that's the only thing we know, hence:
![R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{3}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{5}R =3 \hspace{3}or\hspace{3}G=3](https://tex.z-dn.net/?f=R%3D%5C%7Bx%5Chspace%7B3%7D%20%3B%5Chspace%7B3%7D1%5Cleq%20x%20%5Cleq%206%20%5C%7D%5Chspace%7B3%7Dx%5Cin%20Z%5Chspace%7B3%7DG%3D%5C%7By%5Chspace%7B3%7D%20%3B%5Chspace%7B3%7D1%5Cleq%20y%5Cleq%206%20%5C%7D%5Chspace%7B3%7Dy%5Cin%20Z%5Chspace%7B5%7DR%20%3D3%20%5Chspace%7B3%7Dor%5Chspace%7B3%7DG%3D3)
D: Similar to B, we only know that the sum of R and G isn't 8:
![R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{3}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{5};\hspace{3}R+G\neq8](https://tex.z-dn.net/?f=R%3D%5C%7Bx%5Chspace%7B3%7D%20%3B%5Chspace%7B3%7D1%5Cleq%20x%20%5Cleq%206%20%5C%7D%5Chspace%7B3%7Dx%5Cin%20Z%5Chspace%7B3%7DG%3D%5C%7By%5Chspace%7B3%7D%20%3B%5Chspace%7B3%7D1%5Cleq%20y%5Cleq%206%20%5C%7D%5Chspace%7B3%7Dy%5Cin%20Z%5Chspace%7B5%7D%3B%5Chspace%7B3%7DR%2BG%5Cneq8)